What's the total angular momentum operator for a system of two particles?

[AxiomOfChoice]

Suppose we're in two dimensions, and both particles have mass 1.

Particle 1's location is given by its polar coordinates [itex](r_1,\theta_1)[/itex]; likewise for Particle 2 [itex](r_2,\theta_2)[/itex].

Is it true that the total angular momentum [itex]\vec{L}[/itex] is just the sum of the individual angular momenta of the particles: [itex]\vec{L} = \vec{L}_1 + \vec{L}_2[/itex]? If that's the case, can you give me the total angular momentum operator [itex]\vec{L}[/itex] as a differential operator?

 

[kanato]

yeah, just add it up: [tex] L_j^z = -i\hbar \left[y_j\partial/\partial{x_j} - {x_j}\partial/\partial{y_j}\right][/tex] where j is the particle index. Keep in mind L_2 does not act on the coordinates for the first particle.

 

[Entropia]

Yes, it is true that the total angular momentum operator \vec{L} is the sum of the individual angular momenta of the particles, \vec{L} = \vec{L}_1 + \vec{L}_2. In two dimensions, the total angular momentum operator can be expressed as a differential operator:

\vec{L} = -i\hbar (r_1\frac{\partial}{\partial r_1} + r_2\frac{\partial}{\partial r_2} + \frac{1}{2}(r_1^2\frac{\partial}{\partial \theta_1} + r_2^2\frac{\partial}{\partial \theta_2}))

This operator takes into account the position and momentum of both particles, as well as their angular positions. The \hbar term represents Planck's constant, which is a fundamental constant in quantum mechanics. This operator can be used to calculate the total angular momentum of the system and can be applied to any wave function describing the system.